Gas: Definition, Law, Formula, Equation, Examples, Questions

The Gaseous State

Gases and their properties play an important role in our daily lives. Our atmosphere consists Of a Mixture Of gases like dioxygen, dinitrogen, carbon dioxide, water vapors, etc. These gases shield us from harmful radiation. Life is supported by the dioxygen in the air that we breathe. Plants need carbon dioxide for photosynthesis.
There are in total eleven elements in the periodic table that exist as gases under normal conditions.

The composition of gases is always from non-metal eg. O₂, N₂, He₂, Cl etc.

Following are the few physical properties of the gaseous state :

(i) The volume and shape of gases are not fixed. These assume the volume and shape of the container.

(ii) The thermal energy of gases >> molecular attraction.

(iii) Gases have infinite expansibility and high compressibility.

(iv) Gases exert pressure equally in all directions

(v) Gases have a much lower density than solids and liquids due to negligible intermolecular forces.

vi) Gas mix evenly with other gases or their mixtures are homogeneous in composition.

Characteristics of Gas

There are certain parameters or measurable properties which are used to describe the characteristics of gases

• Volume (V): The volume of the container is the volume of the gas sample as gases occupy the entire space available to them.
  Pressure (P): Pressure Of the gas is the force exerted by the gas per unit area on the walls of the container in all directions.
  $\begin{aligned} & \text { S.I unit }=\text { pascal }(\mathrm{Pa}) \\ & 1 \mathrm{~Pa}=1 \mathrm{Nm}^{-2} \\ & 1 \mathrm{~atm}=1.013 \times 10^5 \mathrm{~Pa} \\ & \text { Conversions } \\ & 1 \mathrm{bar}=10^5 \mathrm{~Pa}=0.987 \mathrm{~atm} \\ & 1 \mathrm{~atm}=760 \mathrm{~mm} \mathrm{Hg} \\ & 1 \mathrm{~atm}=760 \text { torr } \\ & 1 \mathrm{~atm}=1.013 \times 10^5 \mathrm{~Pa}\end{aligned}$
  Temperature: It is the measure of hotness of the system and energy of the system.
• $\begin{aligned} & \text { S.I unit }=\mathrm{m}^3 \\ & \text { C.G.S. unit }=\mathrm{cm}^3 \\ & \text { Commonly used unit }=\mathrm{L} \\ & 1 \mathrm{~L}=1000 \mathrm{~mL} \\ & 1 \mathrm{~mL}=10^{-3} \mathrm{~L} \\ & 1 \mathrm{~m}^3=10^3 \mathrm{dm}^3 \\ & 1 \mathrm{~m}^3=10^3 \mathrm{~L} \\ & 1 \mathrm{~m}^3=10^6 \mathrm{~cm}^3 \\ & 1 \mathrm{~m}^3=10^6 \mathrm{~mL}\end{aligned}$
• $
  \begin{aligned}
  & \text { S.I unit }=\text { kelvin }(\mathrm{K}) \\
  & \mathrm{K}={ }^o \mathrm{C}+273
  \end{aligned}
  $
  ${ }^{\circ} \mathrm{C} \rightarrow$ centigrade degree or Celsius degree

Mass: The mass of a gas can be determined by weighing the container in which the gas is enclosed and again weighing the container after removing the gas. The mass of the gas is related to the number of moles of the gas i.e.Moles of gas $(\mathrm{n})=\frac{\text { Mass in grams }}{\text { Molar mass }}=\frac{\mathrm{m}}{\mathrm{M}}$

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Some Solved Examples

Example 1:
A gas occupies a volume of 500 mL at 27°C and 1 atm pressure. What will be its volume at 47°C and 1 atm pressure?

Solution:
Given:
- Initial volume (V₁) = 500 mL
- Initial temperature (T₁) = 27°C = 27 + 273 = 300 K
- Final temperature (T₂) = 47°C = 47 + 273 = 320 K
- Pressure remains constant at 1 atm

Using the formula: V₁/T₁ = V₂/T₂
V₂ = (V₁ × T₂) / T₁
V₂ = (500 mL × 320 K) / 300 K
V₂ = 533.33 mL

Therefore, the volume of the gas at 47°C and 1 atm pressure is 533.33 mL.

Example 2:
A gas occupies a volume of 2 L at 27°C and 1 atm pressure. What will be its volume at 47°C and 0.5 atm pressure?

Solution:
Given:
- Initial volume (V₁) = 2 L
- Initial temperature (T₁) = 27°C = 27 + 273 = 300 K
- Final temperature (T₂) = 47°C = 47 + 273 = 320 K
- Initial pressure (P₁) = 1 atm
- Final pressure (P₂) = 0.5 atm

Using the formula: (V₁ × P₁) / T₁ = (V₂ × P₂) / T₂
V₂ = (V₁ × P₁ × T₂) / (T₁ × P₂)
V₂ = (2 L × 1 atm × 320 K) / (300 K × 0.5 atm)
V₂ = 4.27 L

Therefore, the volume of the gas at 47°C and 0.5 atm pressure is 4.27 L.

Example 3:
A gas occupies a volume of 1 L at 27°C and 1 atm pressure. How many moles of the gas are present?

Solution:
Given:
- Volume (V) = 1 L = 1000 mL
- Temperature (T) = 27°C = 27 + 273 = 300 K
- Pressure (P) = 1 atm

Using the ideal gas equation: PV = nRT
Where:
- R = 0.082057 L⋅atm⋅mol⁻¹⋅K⁻¹

n = (P × V) / (R × T)
n = (1 atm × 1000 mL) / (0.082057 L⋅atm⋅mol⁻¹⋅K⁻¹ × 300 K)
n = 40.82 mol

Therefore, there are 40.82 moles of the gas present.

Example 4:
A gas occupies a volume of 2 L at 27°C and 1 atm pressure. How many molecules of the gas are present?

Solution:
Given:
- Volume (V) = 2 L = 2000 mL
- Temperature (T) = 27°C = 27 + 273 = 300 K
- Pressure (P) = 1 atm

Using the ideal gas equation: PV = nRT
Where:
- R = 0.082057 L⋅atm⋅mol⁻¹⋅K⁻¹

n = (P × V) / (R × T)
n = (1 atm × 2000 mL) / (0.082057 L⋅atm⋅mol⁻¹⋅K⁻¹ × 300 K)
n = 81.64 mol

1 mol = 6.022 × 10²³ molecules (Avogadro's number)

Number of molecules = n × 6.022 × 10²³
Number of molecules = 81.64 mol × 6.022 × 10²³ molecules/mol
Number of molecules = 4.92 × 10²⁵ molecules

Therefore, there are approximately 4.92 × 10²⁵ molecules of the gas present.

Example 5:
A gas occupies a volume of 3 L at 27°C and 1 atm pressure. What will be its volume at 47°C and 2 atm pressure?

Solution:
Given:
- Initial volume (V₁) = 3 L
- Initial temperature (T₁) = 27°C = 27 + 273 = 300 K
- Final temperature (T₂) = 47°C = 47 + 273 = 320 K
- Initial pressure (P₁) = 1 atm
- Final pressure (P₂) = 2 atm

Using the formula: (V₁ × P₁) / T₁ = (V₂ × P₂) / T₂
V₂ = (V₁ × P₁ × T₂) / (T₁ × P₂)
V₂ = (3 L × 1 atm × 320 K) / (300 K × 2 atm)
V₂ = 3.2 L

Therefore, the volume of the gas at 47°C and 2 atm pressure is 3.2 L.

Summary

Gas molecules travel very fast and far apart, which causes intermolecular forces to become negligible. These forces that keep them moving so fast, coupled with the spacing, give gases a very low density. Gases is that they are highly compressible. Whereas solids and liquids cannot so easily be compressed to exist in smaller volumes, this certainly is not the case for pressurized gases.